Question

if the diagonal of a parallelogram are equal then show that show that it is a rectangle​

Answer 1

To prove: that given parallelogram is a rectangle.

$\rule{70mm}{2pt}$

Step-by-step explanation:

Given that the diagonal of a parallelogram are equal. We need to prove that given parallelogram is a rectangle. Means we need to prove that one of the interior angle is of 90°.

Assumption: Let's say that ABCD is a parallelogram and BD & AC meet each other at point O.

(Refer the attachment for figure)

Now,

In ∆ABC and ∆DCB

BC = BC (common side in both triangles)

AB = DC (opposite sides are equal)

AC = DB (diagonals are equal)

By SSS

∆ABC ≅ ∆DCB

∠ABC = ∠DCB (By CPCT)

AB || DC and BC is transversal. So,

∠ABC + ∠DCB = 180° (interior angles on the same side of the transversal are supplementary)

∠ABC + ∠ABC = 180° (As ∠ABC = ∠DCB)

2∠ABC = 180°

∠ABC = 90°

Since, the opposite angles of a parallelogram are equal. So, all angles of parallelogram are of 90°. Therefore, ABCD is a rectangle as one of it's interior angle is of 90°.

Hence, proved.